I have this matrix in MATLAB:
x = [NaN -2 -1 0 1 2;
1 0.21 0.15 0.34 0.11 0.32;
2 0.14 0.10 0.16 0.31 0.11];
The first row represents the location of the values following X coordinates.
I shift the first row by -0.63, so x becomes:
New_x = [NaN -2.63 -1.63 -0.63 0.37 1.37;
1 0.21 0.15 0.34 0.11 0.32;
2 0.14 0.10 0.16 0.31 0.11];
How can I use interpolation to get the values at specific coordinates of the New_x matrix that we have in the x matrix? ([-2 -1 0 1 2] points)
New_xInterp = [NaN -2.63 .. -2 .. -1.63 .. -1 .. -0.63 .. 0 .. 0.37 .. 1 .. 1.37 .. 2;
1 0.21 .. ? .. 0.15 .. ? .. 0.34 .. ? .. 0.11 .. ? .. 0.32 .. ?;
2 0.14 .. ? .. 0.10 .. ? .. 0.16 .. ? .. 0.31 .. ? .. 0.11 .. ?];
I want to get the '?' values. I tried to use interp2 function but I don't know which step or 2^k-1 interpolated points between coordinates values I have to have in order to get the points like -2, -1, 0, 1, 2.
Thanks !
Since you do not have 2D data, you are only interpolating on one dimension, you only need the function interp1.
This function can work on vector or matrices if necessary, but it require a slight reorganisation of your data.
%% Input
M = [NaN -2 -1 0 1 2;
1 0.21 0.15 0.34 0.11 0.32;
2 0.14 0.10 0.16 0.31 0.11];
%% Demultiplex inputs
x = M(1,2:end).' ; % extract X values, reorder in column
y = M(2:end,2:end).' ; % extract Y values, reorder in columns
%% Interpolate
xn = sort( [x-0.63 ; x] ) ; % Generate the new_x target values
yn = interp1( x-0.63 , y , xn ,'linear','extrap') ; % Interpolate the full matrix in one go
At this point you have your new xn and yn values in columns:
xn= yn=
-2.63 0.21 0.14
-2 0.1722 0.1148
-1.63 0.15 0.1
-1 0.2697 0.1378
-0.63 0.34 0.16
0 0.1951 0.2545
0.37 0.11 0.31
1 0.2423 0.184
1.37 0.32 0.11
2 0.4523 -0.016
I would keep them like that if you have more operations to do on them later on. However, if you want it back into the format you had at the beginning, we can simply rebuild the new full matrix:
%% Rebuild global matrix
Mout = [ M(:,1) , [xn.' ; yn.'] ]
Mout =
NaN -2.63 -2 -1.63 -1 -0.63 0 0.37 1 1.37 2
1 0.21 0.1722 0.15 0.2697 0.34 0.1951 0.11 0.2423 0.32 0.4523
2 0.14 0.1148 0.1 0.1378 0.16 0.2545 0.31 0.184 0.11 -0.016
Maybe you can try interp1 + arrayfun like below
r = sort([x(1,2:end),New_x(1,2:end)]);
New_xInterp = [New_x(:,1),cell2mat(arrayfun(#(k) interp1(New_x(1,2:end),New_x(k,2:end),r),1:size(New_x,1),'UniformOutput',false).')];
which gives
New_xInterp =
NaN -2.63000 -2.00000 -1.63000 -1.00000 -0.63000 0.00000 0.37000 1.00000 1.37000 NA
1.00000 0.21000 0.17220 0.15000 0.26970 0.34000 0.19510 0.11000 0.24230 0.32000 NA
2.00000 0.14000 0.11480 0.10000 0.13780 0.16000 0.25450 0.31000 0.18400 0.11000 NA
The code above used linear interpolation. If you want other options, you can type help interp1 to see more.
Related
I am trying to optimise the running time of my code by getting rid of some for loops. However, I have a variable that is incremented in each iteration in which sometimes the index is repeated. I provide here a minimal example:
a = [1 4 2 2 1 3 4 2 3 1]
b = [0.5 0.2 0.3 0.4 0.1 0.05 0.7 0.3 0.55 0.8]
c = [3 5 7 9]
for i = 1:10
c(a(i)) = c(a(i)) + b(i)
end
Ideally, I would like to compute it by writting:
c(a) = c(a) + b
but obviously it would not give me the same results since I have to recalculate the value for the same index several times so this way to vectorise it would not work.
Also, I am working in Matlab or Octave in case that this is important.
Thank you very much for any help, I am not sure that it is possible to be vectorise.
Edit: thank you very much for your answers so far. I have discovered accumarray, which I did not know before and also understood why changing the for loop between Matlab and Octave was giving me such different times. I also understood my problem better. I gave a too simple example which I thought I could extend, however, what if b was a matrix?
(Let's forget about c at the moment):
a = [1 4 2 2 1 3 4 2 3 1]
b =[0.69 -0.41 -0.13 -0.13 -0.42 -0.14 -0.23 -0.17 0.22 -0.24;
0.34 -0.39 -0.36 0.68 -0.66 -0.19 -0.58 0.78 -0.23 0.25;
-0.68 -0.54 0.76 -0.58 0.24 -0.23 -0.44 0.09 0.69 -0.41;
0.11 -0.14 0.32 0.65 0.26 0.82 0.32 0.29 -0.21 -0.13;
-0.94 -0.15 -0.41 -0.56 0.15 0.09 0.38 0.58 0.72 0.45;
0.22 -0.59 -0.11 -0.17 0.52 0.13 -0.51 0.28 0.15 0.19;
0.18 -0.15 0.38 -0.29 -0.87 0.14 -0.13 0.23 -0.92 -0.21;
0.79 -0.35 0.45 -0.28 -0.13 0.95 -0.45 0.35 -0.25 -0.61;
-0.42 0.76 0.15 0.99 -0.84 -0.03 0.27 0.09 0.57 0.64;
0.59 0.82 -0.39 0.13 -0.15 -0.71 -0.84 -0.43 0.93 -0.74]
I understood now that what I would be doing is rowSum per group, and given that I am using Octave I cannot use "splitapply". I tried to generalise your answers, but accumarray would not work for matrices and also I could not generalise #rahnema1 solution. The desired output would be:
[0.34 0.26 -0.93 -0.56 -0.42 -0.76 -0.69 -0.02 1.87 -0.53;
0.22 -1.03 1.53 -0.21 0.37 1.54 -0.57 0.73 0.23 -1.15;
-0.20 0.17 0.04 0.82 -0.32 0.10 -0.24 0.37 0.72 0.83;
0.52 -0.54 0.02 0.39 -1.53 -0.05 -0.71 1.01 -1.15 0.04]
that is "equivalent" to
[sum(b([1 5 10],:))
sum(b([3 4 8],:))
sum(b([6 9],:))
sum(b([2 7],:))]
Thank you very much, If you think I should include this in another question instead of adding the edit I will do so.
Original question
It can be done with accumarray:
a = [1 4 2 2 1 3 4 2 3 1];
b = [0.5 0.2 0.3 0.4 0.1 0.05 0.7 0.3 0.55 0.8];
c = [3 5 7 9];
c(:) = c(:) + accumarray(a(:), b(:));
This sums the values from b in groups defined by a, and adds that to the original c.
Edited question
If b is a matrix, you can use
full(sparse(repmat(a, 1, size(b,1)), repelem(1:size(b,2), size(b,1)), b))
or
accumarray([repmat(a, 1, size(b,1)).' repelem(1:size(b,2), size(b,1)).'], b(:))
Matrix multiplication and implicit expansion and can be used (Octave):
nc = numel(c);
c += b * (1:nc == a.');
For input of large size it may be more memory efficient to use sparse matrix:
nc = numel(c);
nb = numel(b);
c += b * sparse(1:nb, a, 1, nb, nc);
Edit: When b is a matrix you can extend this solution as:
nc = numel(c);
na = numel(a);
out = sparse(a, 1:na, 1, nc, na) * b;
I have a couple of matrices (1800 x 27) that represent subjects and their recordings (3 minutes equivalent for each of 27 subjects). Each column represents a subject.
I need to do intercorrelation between subjects, let's say to correlate F to G, G to H, and H to F for all 27 subjects.
I use CORR command corr(B) where B is a matrix and it returns the next example:
1 0.07 -0.05 0.10 0.04 0.12
0.07 1 -0.02 -0.08 0.17 0.03
-0.05 -0.02 1 0.04 0.16 0.13
0.10 -0.08 0.04 1 -0.04 0.34
0.04 0.18 0.16 -0.04 1 0.13
How can I adjust the code to exclude self-correlation (eg F to F) so I won't get "1" numerals?
(it's present in each row/column)
I have to perform some transformations afterwards, like Fisher Z-Transformation, which returns inf for each "1", and as result, I can't use further calculations.
I am able to plot a trisurf chart, but surf does not work.
What am I doing wrong?
pkg load statistics;
figure (1,'name','Matrix Map');
colormap('hot');
t = dlmread('C:\Map3D.csv');
tx =t(:,1);ty=t(:,2);tz=t(:,3);
tri = delaunay(tx,ty);
handle = surf(tx,ty,tz); #This does NOT work
#handle = trisurf(tri,tx,ty,tz); #This does work
`error: surface: rows (Z) must be the same as length (Y) and columns (Z) must be the same as length
(X)
My data is in a CSV (commas not shown here)
1 2 -0.32
2 2 0.33
3 2 0.39
4 2 0.09
5 2 0.14
1 2.5 -0.19
2 2.5 0.13
3 2.5 0.15
4 2.5 0.24
5 2.5 0.33
1 3 0.06
2 3 0.44
3 3 0.36
4 3 0.45
5 3 0.51
1 3.5 0.72
2 3.5 0.79
3 3.5 0.98
4 3.5 0.47
5 3.5 0.55
1 4 0.61
2 4 0.13
3 4 0.44
4 4 0.47
5 4 0.58
1 4.5 0.85
surf error message is different in Matlab or in Octave.
Error message from Matlab:
Z must be a matrix, not a scalar or vector.
The problem is pretty clear here since you specified Z (for you tz) as a vector.
Error message from Octave:
surface: rows (Z) must be the same as length (Y) and columns (Z) must be the same as length (X)
You are wrong here since on your example, columns (Z) = 1, but length (X) = 26, so here is the mistake.
One of the consequences of that is that with surf you cannot have "holes" or undefined points on your grid. On your case you have a X-grid from 1 to 5 and a Y-grid from 2 to 4.5 but point of coordinate (2, 4.5) is not defined.
#Luis Mendo, Matlab and Octave do allow the prototype surf(matrix_x, matrix_y, matrix_z) but the third argument matrix_z still have to be a matrix (not a scalar or vector). Apparently, a matrix of only one line or column is not considered as a matrix.
To solve the issue, I suggest something like:
tx = 1:5; % tx is a vector of length 5
ty = 2:0.5:4.5; % ty is a vector of length 6
tz = [-0.32 0.33 0.39 0.09 0.14;
-0.19 0.13 0.15 0.24 0.33;
0.06 0.44 0.36 0.45 0.51;
0.72 0.79 0.98 0.47 0.55;
0.61 0.13 0.44 0.47 0.58;
0.85 0. 0. 0. 0.]; % tz is a matrix of size 6*5
surf(tx,ty,tz);
Note that I had to invent some values at the points where your grid was not defined, I put 0. but you can change it with the value you prefer.
I have astraightline with the following data
xinter=[1.13 1.36 1.62 1.81 2.00 2.30 2.61 2.83 3.05 3.39]
yinter=[0.10 0.25 0.40 0.50 0.60 0.75 0.90 1.00 1.10 1.25]
and I want to find the intersection with a result of an interpolated data
of a curve such as below
a50= [0.77 0.73 0.77 0.85 0.91 0.97 1.05 1.23 1.43 1.53 1.62 1.71 1.89 2.12 2.42];
a25= [0.51 0.60 0.70 0.80 0.85 0.90 0.96 1.09 1.23 1.30 1.36 1.41 1.53 1.67];
vel25=[0.43 0.35 0.30 0.27 0.25 0.24 0.22 0.21 0.22 0.24 0.25 0.27 0.30 0.35];
vel50=[0.68 0.57 0.49 0.43 0.40 0.38 0.36 0.34 0.36 0.38 0.40 0.43 0.49 0.57 0.68 ];
% back up original data, just for final plot
bkp_a50 = a50 ; bkp_vel50 = vel50 ;
% make second x vector monotonic
istart = find( diff(a50)>0 , 1 , 'first') ;
a50(1:istart-1) = [] ;
vel50(1:istart-1) = [] ;
% prepare a 3rd dimension vector (from 25 to 50)
T = [repmat(25,size(a25)) ; repmat(40,size(a50)) ] ;
% merge all observations together
A = [ a25 ; a50] ;
V = [vel25 ; vel50] ;
% find the minimum domain on which data can be interpolated
% (anything outside of that will return NaN)
Astart = max( [min(a25) min(a50)] ) ;
Astop = min( [max(a25) max(a50)] ) ;
% use the function 'griddata'
[TI,AI] = meshgrid( 25:40 , linspace(Astart,Astop,10) ) ;
VI = griddata(T,A,V,TI,AI) ;
% plot all the intermediate curves
%plot(AI,VI)
hold on
% the original curves
%plot(a25,vel25,'--k','linewidth',2)
%plot(bkp_a50,bkp_vel50,'--k','linewidth',2)
% Highlight the curve at T = 30 ;
c30 = find( TI(1,:) == 40 ) ;
plot(AI(:,c30),VI(:,c30),'--r','linewidth',2)
xinter=[1.13 1.36 1.62 1.81 2.00 2.30 2.61 2.83 3.05 3.39]
yinter=[0.10 0.25 0.40 0.50 0.60 0.75 0.90 1.00 1.10 1.25]
x1inter=(AI(:,c30))';
y1inter=(VI(:,c30))';
yy2 = interp1(xinter, yinter, x1inter,'spline')
plot(xinter,yinter, '--k','linewidth',2)
idx = find((y1inter - yy2) < eps, 1); %// Index of coordinate in array
px = x1inter(idx)
py = y1inter(idx)
plot(px, py, 'ro', 'MarkerSize', 18)
But there is an error in the result when I modify x1inter
You can use piecewise polynomial curvefitting and the fzero function to find the intersection point:
pp1 = pchip(xinter,yinter); % Curve 1
pp2 = pchip(AI(:,c30),VI(:,c30)); % Curve 2
fun = #(x) ppval(pp1,x) - ppval(pp2,x); % Curve to evaluate
xzero = fzero(fun,mean(xinter)) % intersection x value
yzero = ppval(pp1,xzero)
plot(xzero, yzero, 'bo', 'MarkerSize', 18)
I have a huge amount of data in MATLAB (350695x5).
An example is like this:
z = [
1.79 0.16 0.16 21.39 21.50
1.83 0.16 0.16 21.39 22.40
1.92 0.16 0.16 21.39 22.00
2.07 0.16 0.16 21.39 22.00
2.36 0.15 0.15 21.39 21.08
2.96 0.13 0.13 21.39 21.04
3.21 0.13 0.13 21.39 23.00
3.72 0.12 0.12 21.39 24.00
3.87 0.11 0.11 21.39 21.39
4.14 0.10 0.10 21.39 22.00
4.14 0.10 0.10 21.39 21.50
4.16 0.10 0.10 21.39 21.39]
and I need to sort it in the following way:
based on 1 column from 1-2, 2-3, 3-4
and find mean values in the range (0-1, 1-2, 2-3, 3-4) for 2,3,4 columns
the result should look like this:
1 0.16 0.16 21.39 21.97
2 0.15 0.15 21.39 21.49
3 0.12 0.12 21.39 22.68
4 0.10 0.10 21.39 21.63
The problem is that I cannot sort it in a proper way.
The part of the solution can be described by
[ii jj] = ndgrid(z(:,1)+1,1:size(z,2)-1) %should sort first column from 0-1,1-2, 2-3, 3-4
z23 = z(:,2:end)
out = [unique(z(:,1)),accumarray([ii(:),jj(:)],z23(:),[],#mean)], %find mean value
Try this:
idx = floor(z(:, 1));
sub = [idx z(:, 2:5)];
[xx, yy] = ndgrid(idx, 1:size(sub, 2));
out = accumarray([xx(:) yy(:)], sub(:), [], #mean)
out =
1.0000 0.1600 0.1600 21.3900 21.9667
2.0000 0.1467 0.1467 21.3900 21.3733
3.0000 0.1200 0.1200 21.3900 22.7967
4.0000 0.1000 0.1000 21.3900 21.6300
Results don't match exactly with yours. I'm not sure I understand exactly what you wanted, but the code I wrote calculates the average on ranges 1 <= x < 2, 2 <= x < 3, and so on.
Use logical indexing to find the values in z where a certain range applies, e.g.:
i01 = (z >= 0) & (z < 1); % Find logical indices
z01 = z(i01); % Get values from 0 up to 1 (but not including 1)
Then, calculation of the mean is easy: mu_z01 = mean(z01);. Of course, the same method can be applied to the other ranges from 1 to 2, 2 to 3, et cetera.